Exponential Growth & Decay

Real growth is usually described as a percent per period, which you convert into a multiplier.

Growing 8% per year: multiply by 1.08 each year, so y = a(1.08)ᵗ.
Decaying 8% per year: multiply by 0.92 each year, so y = a(0.92)ᵗ.

The multiplier is 1 + r for growth and 1 − r for decay, with r as a decimal.

A car worth $24,000 losing 15% per year: V = 24000(0.85)ᵗ. After 3 years, V = 24000(0.614) ≈ $14,739.

Note it never reaches zero. Each year removes 15% of a smaller amount, which is exactly how depreciation and radioactive decay behave.

The models A = a(1 + r)ᵗ and A = a(1 − r)ᵗ express a constant proportional change per period. Compound interest generalises this to n periods per year, A = P(1 + r/n)^{nt}, and as n → ∞ this converges to continuous compounding, A = Pe^{rt}, with e ≈ 2.71828.

Half-life problems are decay expressed by the time constant rather than the rate: A = a(1/2)^{t/h}, where h is the half-life. Doubling time works identically with base 2. Converting between rate form and half-life form requires logarithms, which is why solving for t is deferred to Algebra 2.

Two modelling cautions. First, r must match the period of t — an annual rate with monthly periods requires r/12. Second, unbounded exponential growth is a poor long-run model for populations, which is why logistic models with a carrying capacity replace it in applied work.